Calorimetry. D (PE). (Products). Burning of a Match. System. Surroundings. (Reactants). Potential energy. Energy released to the surrounding as heat. Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 293. Endothermic Reaction Reactant + Energy Product. Surroundings.
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
(Products)
Burning of a MatchSystem
Surroundings
(Reactants)
Potential energy
Energy released to the surrounding as heat
Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 293
Reactant + Energy Product
Surroundings
System
Conservation of Energy in a Chemical ReactionIn this example, the energy of the reactants and products increases,
while the energy of the surroundings decreases.
In every case, however, the total energy does not change.
Surroundings
Energy
System
Before
reaction
After
reaction
Myers, Oldham, Tocci, Chemistry, 2004, page 41
Reactant Product + Energy
Surroundings
System
Conservation of Energy in a Chemical ReactionIn this example, the energy of the reactants and products decreases,
while the energy of the surroundings increases.
In every case, however, the total energy does not change.
Surroundings
System
Energy
Before
reaction
After
reaction
Myers, Oldham, Tocci, Chemistry, 2004, page 41
H2O(l) H2O(s) + heat
melting
freezing
System
Direction of Heat FlowSurroundings
EXOthermic
qsys < 0
ENDOthermic
qsys > 0
System
Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 207
Food joules/grams calories/gram Calories/gram
Protein 17 000 4000 4
Fat 38 000 9000 9
Carbohydrates 17 000 4000 4
1calories = 4.184 joules
1000 calories = 1 Calorie
"science" "food"
Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 51
Experimental Determination of Specific Heat of a Metal
Typical apparatus used in this activity include a boiler (such as large glass beaker), a heat source (Bunsen burner or hot plate), a stand or tripod for
the boiler, a calorimeter, thermometers, samples (typically samples of copper, aluminum, zinc, tin, or lead), tongs (or forceps or string) to handle samples, and a balance.
Styrofoam
cover
Styrofoam
cups
Stirrer
A Coffee Cup CalorimeterZumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 302
thermometer
ignition
wires
stirrer
magnifying
eyepiece
insulating
jacket
air space
bucket
heater
crucible
water
ignition coil
sample
steel bomb
1997 Encyclopedia Britanica, Inc.
Boiling  PE
Liquid  KE
Melting  PE
Solid  KE
Heating Curves140
120
100
80
60
40
Temperature (oC)
20
0
20
40
60
80
100
Time
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
A small piece of ice which lived in a test tube fell in love with a Bunsen burner.
“Bunsen! My flame! I melt whenever I see you” said the ice.
The Bunsen burner replied” “It’s just a phase you’re going through”.
F
140
120
100
80
60
40
20
0
20
40
60
80
100
Heat= m x Cvap
Cv = 2256 J/g
E
D
BP
Heat= m x Cfus
Cf = 333 J/g
Heat= m x DT x Cp, gas
Cp (steam) = 2.042 J/goC
Heat= m x DT x Cp, liquid
Temperature (oC)
Cp = 4.184 J/goC
B
MP
C
Heat= m x DT x Cp, solid
A B warm ice
B C melt ice (solid liquid)
C D warm water
D E boil water (liquid gas)
E D condense steam (gas liquid)
E F superheat steam
Cp (ice) = 2.077 J/goC
A
Heat
140
DH = mol xDHvap
DH = mol xDHfus
120
100
80
Heat = mass xDt x Cp, gas
60
40
Temperature (oC)
20
Heat = mass xDt x Cp, liquid
0
20
40
60
Heat = mass xDt x Cp, solid
80
100
Time
Thin metal wall
Insulated box
Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 291
Hot water Cold Water
90 oC 10 oC
Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 291
Water
(50 oC)
Water
(50 oC)
Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 291
Surroundings
Final
Temperature
Block “B”
Block “A”
SYSTEM
20 g (20oC)
30oC
20 g (40oC)
Al
Al
m = 20 g
T = 40oC
m = 20 g
T = 20oC
What will be the final temperature
of the system ?
a) 60oC b) 30oC c) 20oC d) ?
Assume NO heat energy is “lost” to the surroundings from the system.
?
Surroundings
Final
Temperature
Block “B”
Block “A”
SYSTEM
20 g (20oC)
30.0oC
20 g (40oC)
10 g (20oC)
33.3oC
20 g (40oC)
Al
Al
m = 20 g
T = 40oC
m = 10 g
T = 20oC
What will be the final temperature
of the system ?
a) 60oC b) 30oC c) 20oC d) ?
Assume NO heat energy is “lost” to the surroundings from the system.
Surroundings
Final
Temperature
Block “B”
Block “A”
SYSTEM
20 g (20oC)
30.0oC
20 g (40oC)
10 g (20oC)
33.3oC
20 g (40oC)
10 g (40oC)
26.7oC
20 g (20oC)
Al
Al
m = 20 g
T = 20oC
m = 10 g
T = 40oC
Assume NO heat energy is “lost” to the surroundings from the system.
Surroundings
Final
Temperature
Block “B”
Block “A”
SYSTEM
20 g (20oC)
30.0oC
20 g (40oC)
10 g (20oC)
33.3oC
20 g (40oC)
10 g (40oC)
26.7oC
20 g (20oC)
H2O
Ag
m = 75 g
T = 25oC
m = 30 g
T = 100oC
Real Final Temperature = 26.6oC
Why?
We’ve been assuming ALL materials
transfer heat equally well.
Water has a specific heat Cp = 4.184 J/goC
Silver has a specific heat Cp = 0.235 J/goC
It requires 4.184 Joules of energy to heat 1 gram of water 1oC
and only 0.235 Joules of energy to heat 1 gram of silver 1oC.
In our situation (silver is “hot” and water is “cold”)…
this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy.
The amount of heat required to raise the temperature of one gram of substance by one degree Celsius.
Calculations involving Specific Heat
OR
cp = Specific Heat
q = Heat lost or gained
T = Temperature change
m = Mass
Specific Heats of Some Common
Substances at 298.15 K
Substance Specific heat J/(g.K)
Water (l) 4.18
Water (s) 2.06
Water (g) 1.87
Ammonia (g) 2.09
Benzene (l) 1.74
Ethanol (l) 2.44
Ethanol (g) 1.42
Aluminum (s) 0.897
Calcium (s) 0.647
Carbon, graphite (s) 0.709
Copper (s) 0.385
Gold (s) 0.129
Iron (s) 0.449
Mercury (l) 0.140
Lead (s) 0.129
Molar Heat of Fusion
The energy that must be absorbed in order to convert one mole of solid to liquid at its melting point.
The energy that must be removed in order to convert one mole of liquid to solid at its freezing point.
Latent Heat of Phase Change #2
Molar Heat of Vaporization
The energy that must be absorbed in order to convert one mole of liquid to gas at its boiling point.
The energy that must be removed in order to convert one mole of gas to liquid at its condensation point.
Problem: The molar heat of fusion of water is 6.009 kJ/mol. How much energy is needed to convert 60 grams of ice at 0C to liquid water at 0C?
Mass
of ice
Molar
Mass of
water
Heat
of
fusion
The amount of heat released or absorbed during a chemical reaction.
Endothermic:
Reactions in which energy is absorbed as the reaction proceeds.
Exothermic:
Reactions in which energy is released as the reaction proceeds.
1 BTU (British Thermal Unit) – amount of heat needed to raise 1 pound of water 1oF.
1 calorie  amount of heat needed to raise 1 gram of water 1oC
1 Calorie = 1000 calories
“food” = “science”
Candy bar
300 Calories = 300,000 calories
English
Joules
1 calorie = 4.184 Joules
Metric = _______
DH = mol xDHvap
DH = mol xDHfus
120
100
80
60
Heat = mass xDt x Cp, gas
40
Temperature (oC)
20
Heat = mass xDt x Cp, liquid
0
20
40
60
Heat = mass xDt x Cp, solid
80
100
Time
Cp(ice) = 2.077 J/g oC
It takes 2.077 Joules to raise 1 gram ice 1oC.
X Joules to raise 10 gram ice 1oC.
(10 g)(2.077 J/g oC) = 20.77 Joules
X Joules to raise 10 gram ice 10oC.
(10oC)(10 g)(2.077 J/g oC) = 207.7 Joules
q = Cp . m .DT
Heat = (specific heat) (mass) (change in temperature)
DH = mol xDHvap
DH = mol xDHfus
120
100
80
60
Heat = mass xDt x Cp, gas
40
Temperature (oC)
20
Heat = mass xDt x Cp, liquid
0
20
40
60
Heat = mass xDt x Cp, solid
80
100
Time
Given Ti= 30oC
Tf = 20oC
q = Cp . m .DT
Heat = (specific heat) (mass) (change in temperature)
q = 207.7 Joules
Fe
T = 20oC
mass = 240 g
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC).
When thermal equilibrium is reached, the system has a temperature of 42oC.
Find the mass of the iron.
mass = ? grams

LOSE heat = GAIN heat
 [(Cp,Fe) (mass) (DT)] = (Cp,H2O) (mass) (DT)
 [(0.4495 J/goC) (X g) (42oC  500oC)] = (4.184 J/goC) (240 g) (42oC  20oC)]
 [(0.4495) (X) (458)] = (4.184) (240 g) (22)
Drop Units:
205.9 X = 22091
X = 107.3 g Fe
Calorimetry Problems 2
question #5
mass = 97 g
Au
T = 15oC
mass = 323 g

LOSE heat = GAIN heat
A 97 g sample of gold at 785oC is dropped into 323 g of water, which has an initial
temperature of 15oC. If gold has a specific heat of 0.129 J/goC, what is the final
temperature of the mixture? Assume that the gold experiences no change in state
of matter.
 [(Cp,Au) (mass) (DT)] = (Cp,H2O) (mass) (DT)
 [(0.129 J/goC) (97 g) (Tf  785oC)] = (4.184 J/goC) (323 g) (Tf  15oC)]
Drop Units:
 [(12.5) (Tf  785oC)] = (1.35 x 103) (Tf  15oC)]
12.5 Tf + 9.82 x 103 = 1.35 x 103 Tf  2.02 x 104
3 x 104 = 1.36 x 103 Tf
Tf = 22.1oC
Calorimetry Problems 2
question #8
T = 13oC
mass = 87 g
mass = 59 g

LOSE heat = GAIN heat
If 59 g of water at 13oC are mixed with 87 g of water at 72oC, find the final temperature
of the system.
 [(Cp,H2O) (mass) (DT)] = (Cp,H2O) (mass) (DT)
 [(4.184 J/goC) (87 g) (Tf  72oC)] = (4.184 J/goC) (59 g) (Tf  13oC)
Drop Units:
 [(364.0) (Tf  72oC)] = (246.8) (Tf  13oC)
364 Tf + 26208 = 246.8 Tf  3208
29416 = 610.8 Tf
Tf = 48.2oC
Calorimetry Problems 2
question #9
DH = mol xDHvap
DH = mol xDHfus
120
100
80
60
Heat = mass xDt x Cp, gas
40
Temperature (oC)
20
Heat = mass xDt x Cp, liquid
0
20
40
T = 11oC
60
mass = 38 g
Heat = mass xDt x Cp, solid
ice
80
100
Time
T = 56oC
mass = 214 g

LOSE heat = GAIN heat
A 38 g sample of ice at 11oC is placed into 214 g of water at 56oC.
Find the system's final temperature.
D
water cools
B
warm water
C
A
melt ice
warm ice
D
A
C
B
 [(Cp,H2O(l)) (mass) (DT)] = (Cp,H2O(s)) (mass) (DT) + (Cf) (mass) + (Cp,H2O(l)) (mass) (DT)
 [(4.184 J/goC)(214 g)(Tf  56oC)] = (2.077 J/goC)(38 g)(11oC) + (333 J/g)(38 g) + (4.184 J/goC)(38 g)(Tf  0oC)
 [(895) (Tf  56oC)] = 868 + 12654 + (159) (Tf)]
 895 Tf + 50141 = 868 + 12654 + 159 Tf
 895 Tf + 50141 = 13522 + 159 Tf
36619 = 1054 Tf
Tf = 34.7oC
Calorimetry Problems 2
question #10
DH = mol xDHvap
DH = mol xDHfus
120
100
80
60
Heat = mass xDt x Cp, gas
40
Temperature (oC)
20
Heat = mass xDt x Cp, liquid
0
20
40
60
Heat = mass xDt x Cp, solid
80
100
Time
1102
1102
(1000 g = 1 kg)
238.4 g
25 g of 116oC steam are bubbled into 0.2384 kg of water at 8oC. Find the final temperature of the system.
 [qA + qB + qC] = qD
 [(Cp,H2O) (mass) (DT)] + (Cv,H2O) (mass) + (Cp,H2O) (mass) (DT) = [(Cp,H2O) (mass) (DT)]
qD = (4.184 J/goC) (238.4 g) (Tf  8oC)
qD =  997Tf  7972
qA = [(Cp,H2O) (mass) (DT)]
qC = [(Cp,H2O) (mass) (DT)]
qB = (Cv,H2O) (mass)
qA = [(2.042 J/goC) (25 g) (100o  116oC)]
qC = [(4.184 J/goC) (25 g) (Tf  100oC)]
qA = (2256 J/g) (25 g)
qA =  816.8 J
qA = 104.5Tf  10450
qA =  56400 J
 [qA + qB + qC] = qD
 [  816.8  56400 + 104.5Tf  10450] = 997Tf  7972
816.8 + 56400  104.5Tf + 10450 = 997Tf  7972
67667  104.5Tf = 997Tf  7979
A
75646 = 1102Tf
C
B
Tf = 68.6oC
D
Calorimetry Problems 2
question #11
mass = 264 g
Pb
A 322 g sample of lead (specific heat = 0.138 J/goC) is placed into 264 g of water at 25oC.
If the system's final temperature is 46oC, what was the initial temperature of the lead?
T = ? oC
mass = 322 g
Pb
Tf = 46oC

LOSE heat = GAIN heat
 [(Cp,Pb) (mass) (DT)] = (Cp,H2O) (mass) (DT)
 [(0.138 J/goC) (322 g) (46oC  Ti)] = (4.184 J/goC) (264 g) (46oC 25oC)]
Drop Units:
 [(44.44) (46oC  Ti)] = (1104.6) (21oC)]
 2044 + 44.44 Ti = 23197
44.44 Ti = 25241
Ti = 568oC
Calorimetry Problems 2
question #12
mass = 68 g
458.2
458.2
A sample of ice at –12oC is placed into 68 g of water at 85oC. If the final temperature
of the system is 24oC, what was the mass of the ice?
T = 12oC
mass = ? g
H2O
ice
Tf = 24oC
GAIN heat =  LOSE heat
qA = [(Cp,H2O) (mass) (DT)]
24.9 m
qA = [(2.077 J/goC) (mass) (12oC)]
[ qA + qB + qC ] =  [(Cp,H2O) (mass) (DT)]
[ qA + qB + qC ] =  [(4.184 J/goC) (68 g) (61oC)]
qB = (Cf,H2O) (mass)
333 m
qB = (333 J/g) (mass)
458.2 m =  17339
qC = [(Cp,H2O) (mass) (DT)]
m = 37.8 g
qC = [(4.184 J/goC) (mass) (24oC)]
100.3 m
qTotal = qA + qB + qC
458.2 m
Calorimetry Problems 2
question #13
Energy
Endothermic ReactionEnergy + Reactants Products
Products
Energy
+DH Endothermic
Reactants
Reaction progress
Calorimetry 2
Specific Heat Values
Calorimetry 2
Specific Heat Values
Keys
http://www.unit5.org/chemistry/Matter.html
Heat Energy Problems
Heat Problems (key)
Heat Energy of Water Problems (Calorimetry) Specific Heat Problems
Heat Energy Problems
Heat Problems (key)
Heat Energy of Water Problems (Calorimetry)Specific Heat Problems
Keys
a b c
http://www.unit5.org/chemistry/Matter.html
H2(g) + ½ O2(g)
DH = +242 kJ
Endothermic
242 kJ
Exothermic
286 kJ
Endothermic
DH = 286 kJ
Exothermic
H2O(g)
Energy
+44 kJ
Endothermic
H2O(l)
H2(g) + 1/2O2(g) H2O(g) + 242 kJ DH = 242 kJ
Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 211
Calculate the enthalpy of formation of carbon dioxide from its elements.
C(g) + 2O(g) CO2(g)
Use the following data:
2O(g) O2(g)DH =  250 kJ
C(s) C(g) DH = +720 kJ
CO2(g) C(s) + O2(g) DH = +390 kJ
2O(g) O2(g)DH =  250 kJ
C(g) C(s) DH =  720 kJ
C(s) + O2(g) CO2(g)DH =  390 kJ
C(g) + 2O(g) CO2(g) DH = 1360 kJ
Smith, Smoot, Himes, pg 141
In football, as in Hess's law, only the initial and final conditions matter.
A team that gains 10 yards on a pass play but has a fiveyard penalty,
has the same net gain as the team that gained only 5 yards.
10 yard pass
5 yard net gain
5 yard penalty
initial position
of ball
final position
of ball